# Evaluate the integral. `int te^sqrt(t) dt`

Asked on by nick-teal

### Textbook Question

Chapter 7, Chapter Review - Problem 22 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

tiburtius | High School Teacher | (Level 2) Educator

Posted on

Let us first rewrite the integral a bit.

`int te^sqrt t dt=int(sqrt t ^3 e^sqrt t)/(sqrt t)dt=`

Now we use substitution `x=sqrt t,\ dx=dt/(2sqrt t)`

` ` `2int x^3e^x dx=`

Now we use partial integration `|[u=x^3,dv=e^xdx],[du=3x^2dx,v=e^x]|`

`2(x^3e^x-3int x^2e^x dx)=`

Again we use partial integration `|[u=x^2,dv=e^xdx],[du=2xdx,v=e^x]|`

`2x^3e^x-6(x^2e^x-2int xe^xdx)=`

Partial integration once more `|[u=x,dv=e^xdx],[du=dx,v=e^x]|`

`2x^3e^x-6x^2e^x+12(xe^x-int e^xdx)=`

`2x^3e^x-6x^2e^x+12xe^x-12e^x+C=`

Now we return the substitution.

`2sqrt t ^3e^sqrt t-6te^sqrt t+12sqrt t e^sqrt t-12e^sqrt t+C=`

`e^sqrt t(2sqrt t^3-6t+12sqrt t-12)+C`

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